Probability

1.Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

1. ¹/₂
2. ²/₅
3. ⁸/₁₅
4. ⁹/₂₀

Answer:Option 1

Explanation:

Solution:

Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

∴ P (E) = \frac{n (E)}{n (S)} = \frac{9}{20}

$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{9}{{20}}$

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2.A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

1. ¹⁰/₂₁
2. ¹¹/₂₁
3. ²/₇
4. ⁵/₇

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Total number of balls \\ = (2 + 3 + 2) = 7. \\ Let S be the sample space . \\ Then, n (S) = Number of ways of \\ drawing 2 balls out of 7 \\ =^{7} C_{2} \\ = \frac{(7 \times 6)}{(2 \times 1)} \\ = 21. \\ Let E = Event of drawing 2 balls, \\ none of which is blue . \\ ∴ n (E) = Number of ways of \\ drawing 2 balls out of (2 + 3) balls \\ =^{5} C_{2} \\ = \frac{(5 \times 4)}{(2 \times 1)} \\ = 10. \\ ∴ P (E) = \frac{n (E)}{n (S)} = \frac{10}{21} . \end{aligned}

$\eqalign{ & {\text{Total}}\,{\text{number}}\,{\text{of}}\,{\text{balls}} \cr & = \left( {2 + 3 + 2} \right) = 7. \cr & {\text{Let}}\,{\text{S}}\,{\text{be}}\,{\text{the}}\,{\text{sample}}\,{\text{space}}{\text{.}} \cr & {\text{Then,}}\,n\left( S \right) = {\text{Number}}\,{\text{of}}\,{\text{ways}}\,{\text{of}}\, \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{drawing}}\,{\text{2}}\,{\text{balls}}\,{\text{out}}\,{\text{of}}\,{\text{7}} \cr & = {}^7{C_2} \cr & = \frac{{\left( {7 \times 6} \right)}}{{\left( {2 \times 1} \right)}} \cr & = 21. \cr & {\text{Let}}\,{\text{E = Event}}\,{\text{of}}\,{\text{drawing}}\,{\text{2}}\,{\text{balls,}}\, \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{none}}\,{\text{of}}\,{\text{which}}\,{\text{is}}\,{\text{blue}}{\text{.}} \cr & \therefore n\left( E \right) = {\text{Number}}\,{\text{of}}\,{\text{ways}}\,{\text{of}}\, \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{drawing}}\,{\text{2}}\,{\text{balls}}\,{\text{out}}\,{\text{of}}\,\left( {{\text{2 + 3}}} \right){\text{balls}} \cr & = {}^5{C_2} \cr & = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} \cr & = 10. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{10}}{{21}}. \cr}$

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3.In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

1. ¹/₃
2. ³/₄
3. ⁷/₁₉
4. ⁸/₂₁

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Total number of balls \\ = (8 + 7 + 6) = 21. \\ Let E \\ = event that the ball drawn is \\ neither red nor green \\ = event that the ball drawn is blue . \\ ∴ n (E) = 7. \\ ∴ P (E) = \frac{n (E)}{n (S)} = \frac{7}{21} = \frac{1}{3} \end{aligned}

$\eqalign{ & {\text{Total}}\,{\text{number}}\,{\text{of}}\,{\text{balls}} \cr & = \left( {8 + 7 + 6} \right) = 21. \cr & {\text{Let}}\,{\text{E}} \cr & {\text{ = event}}\,{\text{that}}\,{\text{the}}\,{\text{ball}}\,{\text{drawn}}\,{\text{is}} \cr & {\text{ }}\,\,\,\,\,\,{\text{neither}}\,{\text{red}}\,{\text{nor}}\,{\text{green}} \cr & {\text{ = event}}\,{\text{that}}\,{\text{the}}\,{\text{ball}}\,{\text{drawn}}\,{\text{is}}\,{\text{blue}}{\text{.}} \cr & \therefore n\left( E \right) = 7. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{7}{{21}} = \frac{1}{3} \cr}$

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4.Three unbiased coins are tossed. What is the probability of getting at most two heads?

1. ³/₄
2. ¹/₄
3. ³/₈
4. ⁷/₈

Answer:Option 1

Explanation:

Solution:

Getting at most Two heads means 0 to 2 but not more than 2.
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

∴ P (E) = \frac{n (E)}{n (S)} = \frac{7}{8}

$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{7}{8}$

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5.Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

1. ¹/₂
2. ³/₄
3. ³/₈
4. ⁵/₁₆

Answer:Option 1

Explanation:

Solution:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

\begin{aligned} ∴ n (E) = 27 \\ ∴ P (E) = \frac{n (E)}{n (S)} \\ = \frac{27}{36} = \frac{3}{4} \end{aligned}

$\eqalign{ & \therefore n\left( E \right) = 27 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{27}}{{36}} = \frac{3}{4} \cr}$

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6.In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

1. ²¹/₄₆
2. ²⁵/₁₁₇
3. ¹/₅₀
4. ³/₂₅

Answer:Option 1

Explanation:

Solution:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25.

\begin{aligned} =^{25} C_{3} \\ = \frac{(25 \times 24 \times 23)}{(3 \times 2 \times 1)} \\ = 2300. \\ n (E) = (^{10} C_{1} \times^{15} C_{2}) \\ = [10 \times \frac{(15 \times 14)}{(2 \times 1)}] \\ = 1050. \\ ∴ P (E) = \frac{n (E)}{n (S)} = \frac{1050}{2300} = \frac{21}{46} . \end{aligned}

$\eqalign{ & = {}^{25}{C_3} \cr & = \frac{{\left( {25 \times 24 \times 23} \right)}}{{\left( {3 \times 2 \times 1} \right)}} \cr & = 2300. \cr & n\left( E \right) = \left( {{}^{10}{C_1} \times {}^{15}{C_2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {10 \times \frac{{\left( {15 \times 14} \right)}}{{\left( {2 \times 1} \right)}}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1050. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}}. \cr}$

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7.In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

1. ¹/₁₀
2. ²/₅
3. ²/₇
4. ⁵/₇

Answer:Option 1

Explanation:

Solution:

\begin{aligned} P (getting a prize) \\ = \frac{10}{10 + 25} \\ = \frac{10}{35} \\ = \frac{2}{7} \end{aligned}

$\eqalign{ & P\left( {{\text{getting}}\,{\text{a}}\,{\text{prize}}} \right) \cr & = \frac{{10}}{{10 + 25}} \cr & = \frac{{10}}{{35}} \cr & = \frac{2}{7} \cr}$

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8.Two dice are tossed. The probability that the total score is a prime number is:

1. ¹/₆
2. ⁵/₁₂
3. ¹/₂
4. ⁷/₉

Answer:Option 1

Explanation:

Solution:

Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.Then
E = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) }

\begin{aligned} ∴ n (E) = 15. \\ ∴ P (E) = \frac{n (E)}{n (S)} \\ = \frac{15}{36} \\ = \frac{5}{12} \end{aligned}

$\eqalign{ & \therefore n\left( E \right) = 15. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{15}}{{36}} \cr & = \frac{5}{{12}} \cr}$

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9.A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:

1. ¹/₁₃
2. ²/₁₃
3. ¹/₂₆
4. ¹/₅₂

Answer:Option 1

Explanation:

Solution:

Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2.

\begin{aligned} ∴ P (E) = \frac{n (E)}{n (S)} \\ = \frac{2}{52} \\ = \frac{1}{26} \end{aligned}

$\eqalign{ & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{2}{{52}} \cr & = \frac{1}{{26}} \cr}$

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10.A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

1. ¹/₂₂
2. ³/₂₂
3. ²/₉₁
4. ²/₇₇

Answer:Option 1

Explanation:

Solution:

Let S be the sample space.
Then, n(S) = number of ways of drawing 3 balls out of 15

\begin{aligned} =^{15} C_{3} \\ = \frac{(15 \times 14 \times 13)}{(3 \times 2 \times 1)} \\ = 455. \\ Let E = event of getting all the 3 red balls \\ ∴ n (E) =^{5} C_{3} =^{5} C_{2} \\ = \frac{(5 \times 4)}{(2 \times 1)} = 10. \\ ∴ P (E) = \frac{n (E)}{n (S)} \\ = \frac{10}{455} \\ = \frac{2}{91} . \end{aligned}

$\eqalign{ & = {}^{15}{C_3} \cr & = \frac{{\left( {15 \times 14 \times 13} \right)}}{{\left( {3 \times 2 \times 1} \right)}} \cr & = 455. \cr & {\text{Let}}\,{\text{E}}\,{\text{ = }}\,{\text{event}}\,{\text{of}}\,{\text{getting}}\,{\text{all}}\,{\text{the}}\,{\text{3}}\,{\text{red}}\,{\text{balls}} \cr & \therefore n\left( E \right) = {}^5{C_3} = {}^5{C_2} \cr & = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} = 10. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{10}}{{455}} \cr & = \frac{2}{{91}}. \cr}$

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